Hello there! I am currently attempting to make improvements to an existing program that is poorly written. Specifically, I am working on a system using an A-B MicroLogix 1200. In the ladder diagram provided, I have a question: If Input B is not activated but Input A is, will the coil still be activated?
The coil value is updated individually for each rung in the ladder logic. On the second rung, the coil is set to false. Because the 1200 PLC updates outputs at the end of each scan, there is no chance for it to become energized, not even for a moment.
Imagine toggling offline in a program undergoing updates and taking on the logic yourself. Start by scanning from top to bottom, left to right. If Input A is false, then Coil A is Low. If Input B is false, Coil A remains Low. Send data to ensure Coil A stays low. In the next scan, if Input A is true, Coil A becomes high. If Input B is false, Coil A stays low. Send data to maintain Coil A low based on the last received information. In the following scan, if Input A is false, Coil A is low. If Input B is true, Coil A becomes high. Send data to set Coil A high according to the last received information. By reading from top to bottom, left to right, the final decision for the coil will be based on the last read state of the logic. While there may be instances where there is a (L) in multiple places, causing intermittent output issues, an OTE is a one-time use, and any instance of the coil before the last one scanned is irrelevant to the final decision-making process.
The OTE coil instruction stands out for its unique functionality - it performs actions both when it is true and when it is false. This differs from most output instructions, which only act when true and remain inactive when false. As pointed out by Cheeseface, the physical outputs are updated at the end of the scan cycle. In cases where two rungs try to control the same output using OTE instructions, the principle of "last one wins" applies. Consequently, Input_B will always determine the status of the physical output Coil_A. The first rung becomes redundant and can be deleted since it has no impact on the physical output. If Input_B is open, the output is disabled, whereas if Input_B is closed, the output is activated. Despite being technically valid in the code, this setup may not function as expected due to the programming logic involved.
Thank you everyone for your help.
Ron Beaufort provides a simple explanation for Output Energizing (OTE) in PLC programming. Essentially, it assigns a value of 1 to the tag when the previous rung is true, and a value of 0 when the rung is false. This action does not physically activate an output. The physical output is only affected when the tag is evaluated during the IO update process. In Micro Logix, the IO update occurs at the end of the ladder scan, but in situations like immediate IO or Control Logix, the IO may be updated asynchronously to the scan. To ensure sequential IO update, it is recommended to map real-world IO to intermediate tags and incorporate these tags into your logic. This practice also facilitates the reuse of code across different PLCs with varying IO configurations.
Hello! Based on what you've shared, if only Input A is being activated but not Input B, then the coil will not get activated if you're following the standard AND logic in ladder diagrams. Each parallel rung behaves like an OR statement, and each series rung behaves like an AND statement. So, unless both inputs on the same rung are activated (input A AND input B), the coil won't trigger. However, you can consider changing the logic to OR if you want the coil to activate when only one of the inputs is active.
Hey! For an A-B MicroLogix 1200 and similar PLCs, the answer depends on how the ladder logic is designed. If the rung conditions for the coil only depend on Input A being TRUE, then yes, the coil will be activated. However, if the conditions are set such that Inputs A and B have AND logic implemented (where both inputs need to be TRUE to activate the coil), the coil won't be activated if only Input A is TRUE and B is not. Hope this clarifies things a bit!
Hey there! In the scenario you've described, if Input B isn't triggered but Input A is, ideally the coil should not activate. The nature of ladder diagrams is such that all conditions in the rung (in this case Inputs A and B) need to be true for the output (the coil) to be activated. However, it really depends on how your circuit is wired, if it's an AND condition, then both inputs need to be active. But, if it's an OR condition, either of the inputs will suffice. Hope this helps!
Great question! In a ladder diagram, if you have a series circuit where Input A and Input B are in series, then the coil will not activate unless both inputs are on. However, if the inputs are wired in parallel, the coil will activate as long as Input A is energized, regardless of the state of Input B. It’s important to double-check your circuit logic to see how those inputs are arranged!
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Answer: 1. Will the coil be activated if Input B is not active but Input A is in an A-B MicroLogix 1200 system? - No, the coil will not be activated if Input B is not active even if Input A is active. Both Input A and Input B need to be active for the coil to be activated in an A-B MicroLogix 1200 system.
Answer: - In ladder logic programming, the logical AND function is represented by connecting two inputs with an AND instruction. This means that for the coil to be activated, both Input A and Input B must be active simultaneously to satisfy the logical AND condition.
Answer: - To achieve this behavior, you can use an OR instruction in the ladder logic programming. By connecting Input A and Input B with an OR instruction before the coil, the coil will be activated if either Input A or Input B is active, providing the desired functionality.
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